Solved CAT Questions (Algebra)  Set 4

Q1. (CAT 2000)
For three distinct real numbers x, y and z, let
f (x, y, z) = min(max(x, y), max(y, z), max(z, x))
g(x, y, z) = max(min(x, y), min(y, z), min(z, x))
h(x, y, z) = max(max(x, y), max(y, z), max(z, x))
j(x, y, z) = min(min(x, y), min(y, z), min(z, x))
m(x, y, z) = max(x, y, z)
n(x, y, z) = min(x, y, z)Which of the following is necessarily greater than 1?
(1) (h(x, y, z) – f(x, y, z))/j(x, y, z)
(2) j(x, y, z)/h(x, y, z)
(3) f(x, y, z)/g(x, y, z)
(4) (f(x, y, z) + h(x, y, z) – g(x, y, z))/j(x, y, z)Which of the following expressions is necessarily equal to 1 ?
(1) (f(x, y, z) – m(x, y, z))/(g(x, y, z) – h(x, y, z))
(2) (m(x, y, z) – f(x, y, z))/(g(x, y, z) – n(x, y, z))
(3) (j(x, y, z) – g(x, y, z))/h(x, y, z)
(4) (f(x, y, z) – h(x, y, z))/f(x, y, z)Which of the following expressions is indeterminate?
(1) (f(x, y, z) – h(x, y, z))/(g(x, y, z) – j(x, y, z))
(2) (f(x, y, z) + h(x, y, z) + g(x, y, z) + j(x, y, z))/(j(x, y, z) + h(x, y, z) – m(x, y, z) – n(x, y, z))
(3) (g(x, y, z) – j(x, y, z))/(f(x, y, z) – h(x, y, z))
(4) (h(x, y, z) – f(x, y, z))/(n(x, y, z) – g(x, y, z))Solution: We need to work with options here. Don't get confused with all the functions and min/max representation. We will put it into a much more simpler way.
For time being consider x > y > z ( say x = 2, y = 3 and z = 5)
First function, f(2, 3, 5) = min(max(2, 3), max(3, 5), max(5, 2)) = min(3, 5, 5) = 3, which is the middle value. (we can get this by observation itself. I wrote the first one just as an explanation)
Similarly g(2, 3, 5)  middle value (3)
h(2, 3, 5)  max value (5)
j(2, 3, 5)  min value (2)
m(2, 3, 5)  max value (5)
n(2, 3, 5)  min value (2)First question  Check in which of the given fractions Numerator > Denominator always. If it helps, you can put values and see the results. Only option 4 satisfies.
Option 4  ( f + h  g) / j = (3 + 5  3) / 2 > 1Second Question  Again check with options.
Option 1  (f  m)/(g  h) = (3  5)/(3  5) = 1.Third question  Option 2, Denominator is zero ( j + h  m  n = 2 + 5  5  2 = 0), hence indeterminate.
Q2. (CAT 2000)
Given below are two graphs made up of straight line segments shown as thick lines. In each case choose the answer as
a. if f(x) = 3 f(–x)
b. if f(x) = –f(–x)
c. if f(x) = f(–x)
d. if 3f(x) = 6f(–x), for x ≥ 0Solution : c and d.
first graph  symmetric graph with respect to y axis. This is a even function, f(x)=f(x)
second graph  this one is a symmetric graph too. But 2 * f(1) = f(1). Answer will be the option where f(x)/f(x) = 2, (option d)Q3. (CAT 2000)
For real numbers x, y, let
f(x, y) = Positive squareroot of (x + y), if (x + y)^0.5 is real
= (x + y)^2, otherwise
g(x, y) = (x + y)^2, if (x + y)^0.5 is real
= –(x + y), otherwise
Which of the following expressions yields a positive value for every pair of nonzero real number (x, y)?
(1) f(x, y) – g(x, y)
(2) f(x, y) – (g(x, y))^2
(3) g(x, y) – (f(x, y))^2
(4) f(x, y) + g(x, y)Solution: f(x, y) is always positive.
g(x, y) is always positive too. (x + y)^2 always positive when (x + y)^0.5 is real
 when (x + y)^0.5 is not real, x + y < 0 so  ( x + y ) is positive
So f(x,y) + g(x,y) is always positive.
Q4. (CAT 2000)
For a real number x,
let f(x) = 1/(1 + x), if x is nonnegative
= 1+ x, if x is negative
f^n(x) = f(f^(n – 1)(x)), n = 2, 3, ....
What is the value of the product, f(2) x f^2(2) x f^3(2) x f^4(2) x f^5(2) ?
(1) 1/3
(2) 3
(3) 1/18
(4) None of thesef(2) = 1/3
f^2(2) = f(f(2)) = f(1/3) = 1/(1 + 1/3) = 3/4
f^3(2) = f(f^2(2)) = f(3/4) = 4/7
f^4(2) = f(f^3(2)) = f(4/7) = 7/11
f^5(2) = f(f^4(2)) = f(7/11) = 11/18
Product = 1/3 x 3/4 x 4/7 x 7/11 x 11/18 = 1/18Q5. (CAT 2000)
For all nonnegative integers x and y, f(x, y) is defined as below
f(0, y) = y + 1
f(x + 1, 0) = f(x, 1)
f(x + 1,y + 1) = f(x, f(x + 1, y))
Then, what is the value of f(1, 2)?
(1) Two
(2) Four
(3) Three
(4) Cannot be determinedTake third equation, f(x + 1, y + 1) = f(x, f(x + 1, y)
We need to find f(1,2) so take x = 0 and y = 1.
f(1, 2) = f(0, f(1, 1))
= f(0, f(0, f (1, 0))
= f(0, f(0, f(0, 1))
= f(0, f(0, 2))
= f(0, 3)
= 4Q6. (CAT 2000)
If the equation x^3– ax^2 + bx – a = 0 has three real roots, then it must be the case that,
a) b = 1
b) b ≠ 1
c) a = 1
d) a ≠ 1Make use of options.
if b = 1, equation reduces to x^3  ax^2 + x  a = 0
x^2 (x  a) + (x  a) = 0
(x  a) (x^2 + 1) = 0
x = a or x^2 = 1
Now we need three REAL solutions, which is not the case here.
So b = 1 cannot be the case, so b should not be equal to one. (b # 1)Q7  Q 10 (CAT 1999)
Answer the questions based on the following information.
In each of the following questions, a pair of graphs F(x) and F1(x) is given. These are composed of straight line segments, shown as solid lines, in the domain x is element of (2, 2).
a. if F1(x) = – F(x)
b. if F1(x) = F(–x)
c. if F1(x) = – F(–x)
d. if none of the above is trueQ7.
F(0) = 0 ; F1(0) = 0
F(2) = 2 ; F1(2) = 2
F(2) = 2 ; F1(2) = 2
None of the above. (F(x) =  F(x)  )Q8.
F1(x) = F(x), so option B
Q9.
F1(x) = F(x), so option B
Q 10.
F1(x) = – F(–x). So Option C